# Area problem - 13 Jan 2016

**Question**

Parallelogram !!ABCD!! formed by the line !!y = hx!!, !!y = hx + 1!!, !!y = kx!! and !!y = kx + 1!!

Then the area of !!ABCD!! ?

!!a)" "|h-k|/(h-k)^2!!

!!b)" "2/|h-k|!!

!!c)" "2/|h+k|!!

!!d)" "1/|h-k|!!

**Solution**

According to the given figure, parallelogram !!ABCD!!

Let equation of line !!AB!! : !!y = hx!!

Equation of line !!BC!!: !!y = kx!!

Equation of line !!CD!!: !!y = hx + 1!!

Equation of line !!DA!!: !!y = kx + 1!!

We need to find the points of intersection of lines in order to get coordinates points.

After simplification we will get,

!!A(1/(h-k), h/(h-k))!!, !!B(0,0)!!, !!C(1/(k-h), k/(k-h))!! and !!D(0,1)!!

Area of parallelogram !!ABCD!! = !!(1/2)|(0,0),(1/(k-h), k/(k-h)),(0,1),(1/(h-k), h/(h-k)),(0,0)|!!

!!= (1/2)|0 + 1/(k-h) - 1/(h-k)|!!

!!= 1/|h-k|!!